Answer:

Explanation:
If the collision is elastic and exactly head-on, then we can use the law of momentum conservation for the motion of the 2 balls
Before the collision

After the collision

So using the law of momentum conservation


We can solve for the speed of ball 1 post collision in terms of others:

Their kinetic energy is also conserved before and after collision


From here we can plug in 






To solve this problem it is necessary to apply the concepts related to the concept of overlap and constructive interference.
For this purpose we have that the constructive interference in waves can be expressed under the function

Where
a = Width of the slit
d = Distance of slit to screen
m = Number of order which represent the number of repetition of the spectrum
Angle between incident rays and scatter planes
At the same time the distance on the screen from the central point, would be

Where y = Represents the distance on the screen from the central point
PART A ) From the previous equation if we arrange to find the angle we have that



PART B) Equation both equations we have


Re-arrange to find a,


Inertia is a force which keeps stationary objects at rest and moving objects in motion at ... False - Pounds is a unit of force commonly used in the British system of ... When a chemistry student places a beaker on a balance and determines it to be ... In this case, an object moving to the right could have a balance of forces if it is ...
The mass of the hoop is the only force which is computed by:F net = 2.8kg*9.81m/s^2 = 27.468 N
the slow masses that must be quicker are the pulley, ring, and the rolling sphere.
The mass correspondent of M the pulley is computed by torque τ = F*R = I*α = I*a/R F = M*a = I*a/R^2 --> M = I/R^2 = 21/2*m*R^2/R^2 = 1/2*m
The mass equal of the rolling sphere is computed by: the sphere revolves around the contact point with the table. So using the proposition of parallel axes, the moment of inertia of the sphere is I = 2/5*mR^2 for spin about the midpoint of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere. I = 7/5*mR^2 M = 7/5*m
the acceleration is then a = F/m = 27.468/(2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2
Answer:
The coefficient of kinetic friction between the puck and the ice is 0.11
Explanation:
Given;
initial speed, u = 9.3 m/s
sliding distance, S = 42 m
From equation of motion we determine the acceleration;
v² = u² + 2as
0 = (9.3)² + (2x42)a
- 84a = 86.49
a = -86.49/84
|a| = 1.0296
= ma
where;
Fk is the frictional force
μk is the coefficient of kinetic friction
N is the normal reaction = mg
μkmg = ma
μkg = a
μk = a/g
where;
g is the gravitational constant = 9.8 m/s²
μk = a/g
μk = 1.0296/9.8
μk = 0.11
Therefore, the coefficient of kinetic friction between the puck and the ice is 0.11